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December 12th, 2017, 07:19 AM   #1
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How to solve this problem involving friction?

The problem is as follows:

In the figure shown, Calculate the Force labeled as $F$ which must be applied to the wooden block $B$ whose mass is $10\textrm{ kg}$ thus it would not slide over the wall. The friction coefficient in the surface is $\frac{1}{3}$ and gravity acceleration is $g=10\,\frac{m}{s^{2}}$

I thought to solve the problem by summing all forces as explained below:

$mg+F\cos37^{o}-\mu\times n=0$

The normal in this case is the same as $F\sin37^{o}$

By plug-in the values in the equation I got to this:

$(10\,kg)(10\frac{m}{s^{2}})+F\cos37^{o}-(\frac{1}{3})\times F\sin37^{o}=0$

$(10\,kg)(10\frac{m}{s^{2}})+F(\frac{3}{5})-(\frac{1}{3})\times F(\frac{3}{5})=0$

$F(\frac{1}{5})(3-1)=-(10\,kg)(10\frac{m}{s^{2}}$

$F(\frac{2}{5})=-100\,kg\times\frac{m}{s^{2}}$

$F=-250\,kg\times\frac{m}{s^{2}}$

However I'm confused as why? the Force is negative.
Attached Images
 g6618.jpg (7.4 KB, 0 views) g5956.jpg (13.3 KB, 0 views)

 December 12th, 2017, 08:26 AM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 239 Thanks: 126 Math Focus: Dynamical systems, analytic function theory, numerics I can't follow your computations at all but you seem to have a few misconceptions which are noticeable. Likely, these are the reason your computation is not intelligible. 1. There is no reason force can't be negative. 2. Force is a vector. 3. The expression $\mu \times n$ is a vector cross product, not multiplication. 4. You can't do normal algebra on vectors. Specifically, you have $F = ma$ but in this expression $F,a$ are both vectors. So you can't "solve" for $m$ as $m = \frac{F}{a}$. The expression on the right is meaningless. I would suggest you forget this particular problem for the moment and revisit the fundamentals of matrix/vector algebra.
December 12th, 2017, 08:58 AM   #3
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Quote:
 Originally Posted by SDK I can't follow your computations at all but you seem to have a few misconceptions which are noticeable. Likely, these are the reason your computation is not intelligible.
I attached a free body diagram in order to explain why I came up to the conclution of the equations stated in my earlier post. I assumed forces pointing in opposite directions would cancel and those pointing to the same direction would sum and the total would be zero. But does this reasoning and the diagram I sketched were wrong?.

Quote:
 Originally Posted by SDK 1. There is no reason force can't be negative. 2. Force is a vector. 3. The expression $\mu \times n$ is a vector cross product, not multiplication. 4. You can't do normal algebra on vectors. Specifically, you have $F = ma$ but in this expression $F,a$ are both vectors. So you can't "solve" for $m$ as $m = \frac{F}{a}$. The expression on the right is meaningless. I would suggest you forget this particular problem for the moment and revisit the fundamentals of matrix/vector algebra.
I'll follow your recommendations but I must say that in my calculations I did included vector decomposition of the Forces acting over the object, therefore the $\sin 37^{o}$ and $\cos37^{o}$ which are next $F$ in the initial equation I posted.

Although I'm aware that this problem can be solved by going on different routes I would prefer to follow the one which avoid the use of calculus and rely on simple algebra.

 December 12th, 2017, 09:55 AM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,691 Thanks: 1350 two cases ... 1. If $F\cos(37) < mg$ then the force of static friction acts upward to maintain equilibrium ... $F\cos(37) + f = mg$ note that $f \le \mu \cdot N$, where $N = F\sin(37)$, so the maximum upward force of friction is $\mu F\sin(37)$ which in turn gives a minimum value for $F$ ... $F_{min}\cos(37) + \mu F_{min}\sin(37) = mg \implies F_{min} = \dfrac{mg}{\cos(37) + \mu\sin(37)}$ 2. If $F\cos(37) > mg$ then the force of static friction acts downward to maintain equilibrium ... $F\cos(37) - f = mg$ in this case, if $f = \mu F\sin(37)$, then $F$ is a maximum ... $F_{max}\cos(37) - \mu F_{max}\sin(37) = mg \implies F_{max} = \dfrac{mg}{\cos(37) - \mu\sin(37)}$ therefore, the range of possible values for $F$ for the system to maintain equilibrium is ... $\dfrac{mg}{\cos(37) + \mu\sin(37)} \le F \le \dfrac{mg}{\cos(37) - \mu\sin(37)}$
 December 12th, 2017, 10:53 AM #5 Senior Member   Joined: Jun 2015 From: England Posts: 712 Thanks: 203 This is where you have to use some Physics common sense, even though it conflicts with the maths. No F can't be negative. Not because you can't apply a negative F (you could) or because the maths is wrong. But simply because that would pull the block away from the wall. What the negative sign is telling you is that you have one of the other forces pointing in the wrong direction. Now the weight (force due to gravity on the mass of the block) is fixed in magnitude and direction and you have that correct. You have n, the normal reaction from the wall onto the block going in the wrong direction however, in your free body diagram. Your n is the force exerted by the block on the wall. But your free body diagram is about forces on the block. Once F is fixed, n is fixed (not variable). Now for the difficult one - the friction. The force of friction is variable and only provides enough force to maintain the balance of equilibrium. This could be anything from zero up to the maximum value of limiting friction. Secondly Friction always acts in the opposite direction from the motion (or proposed motion). Now there are two situations. The force F holds the block against the wall with its horizontal component. Thus this equals the reaction between them. But F may either tend to push the block up the wall (in which case friction will act downwards) Or it will not be strong enough to hold the block up and the blcok will tend to slip down the wall. In this case friction will act upwards and combine with (add to) the vertical component of F to support the block in place. I will not repeat the admirable maths that skeeter has already offered, This ramble was by way of explanation.
December 13th, 2017, 02:02 AM   #6
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Quote:
 Originally Posted by studiot This is where you have to use some Physics common sense, even though it conflicts with the maths. No F can't be negative. Not because you can't apply a negative F (you could) or because the maths is wrong. But simply because that would pull the block away from the wall. What the negative sign is telling you is that you have one of the other forces pointing in the wrong direction.
Checked! It's clear to me now and it makes sense as you mentioned a negative force would meant that the force is pulling in the other direction.

Quote:
 Originally Posted by studiot Now the weight (force due to gravity on the mass of the block) is fixed in magnitude and direction and you have that correct.
Thanks, at least I didn't got that wrong.

Quote:
 Originally Posted by studiot You have n, the normal reaction from the wall onto the block going in the wrong direction however, in your free body diagram. Your n is the force exerted by the block on the wall. But your free body diagram is about forces on the block. Once F is fixed, n is fixed (not variable).
I had a mistake in the way how I sketched the normal force. It seems that the normal is pointing in the direction against the surface which the block is placed. Is this because the definition of normal, always making a $90^{o}$ to the contact surface?

I also notice that there was an error in vector decomposition. I'm comparing both version in the sketch attached.

Quote:
 Originally Posted by studiot Now for the difficult one - the friction. The force of friction is variable and only provides enough force to maintain the balance of equilibrium. This could be anything from zero up to the maximum value of limiting friction. Secondly Friction always acts in the opposite direction from the motion (or proposed motion). Now there are two situations. The force F holds the block against the wall with its horizontal component. Thus this equals the reaction between them. But F may either tend to push the block up the wall (in which case friction will act downwards) Or it will not be strong enough to hold the block up and the blcok will tend to slip down the wall. In this case friction will act upwards and combine with (add to) the vertical component of F to support the block in place. I will not repeat the admirable maths that skeeter has already offered, This ramble was by way of explanation.
Before continuing into the details that you explained. I'm confused. Why this problem is about static friction and not kinetic friction?. On which case kinetic friction would be applied?. By extension to that question, does the coefficient of kinetic friction is also between an interval such as static friction?. In my physics book does say however the numeric value between both is the same. Does that mean kinetic friction is between an interval?. Can you explain me what does make them different?

I'm confused about the force F which you say is pushing the block which causes the friction to act downwards. Is this force a maximum or a minimum. Why? The same applies to the force which causes the friction to act upwards. Is the latter a minimum or a maximum? If you could make a drawing explaining this could be perfect.

Sorry if my questions do seem silly but I'm in the process of clearing my ideas.

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